I am trying to solve the recurrence:
$$
a_{n+2} = \sqrt{a_{n+1}\cdot a_{n}}
$$
but here is a problem for me. After few steps I have this:
$$
a_n^2 = a_{n-1}\cdot a_{n-2}
$$
and I don't now what to do further. I can solve a recurrence like that
$$
a_{n+2} + a_{n+1} - a_n = 5 \cdot 2^n,
$$
but I can't find any information about this case (when I have some degree or
square root in a recurrence).

Where did you get $k$? If $b_n=\log a_n$ then $e^{b_n}=a_n$.
–
Thomas AndrewsNov 28 '12 at 19:18

:) Now I've the same question. Where did you get that $ e $ ?
–
Buga1234Nov 28 '12 at 19:24

1

You don't need to use natural logarithms, you can use any logarithm base you want with this problem. If you define $b_n=\log_{10} a_n$ then you know that $10^{b_n}=a_n$.
–
Thomas AndrewsNov 28 '12 at 19:58